Question

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the past 30 days, with a tolerable sampling error (E) of 0.03 and a confidence level of 95%. If prior secondary data indicated that 25% of all adults had used the Internet for such a purpose, what is the required sample size for the new study

Answer 1

The required sample size for the new study is 801.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

25% of all adults had used the Internet for such a purpose

This means that
`\pi = 0.25`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

What is the required sample size for the new study?

This is n for which M = 0.03. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.25*0.75))/(n)}`

`0.03√(n) = 1.96√(0.25*0.75)`

`√(n) = (1.96√(0.25*0.75))/(0.03)`

`(√(n))^2 = ((1.96√(0.25*0.75))/(0.03))^2`

`n = 800.3`

Rounding up:

The required sample size for the new study is 801.